\rm\color{red}{\large \int_0^\infty (x^(\frac\pi5-1))/(1+x^(2\pi)) \mathrm dx} \\​

Question

`\rm\color{red}{\large \int_0^\infty (x^(\frac\pi5-1))/(1+x^(2\pi)) \mathrm dx} \\`​

Answer 1

Substitute
`y=x^(\pi/5)` and
`dy=\frac\pi5 x^(\pi/5 - 1) \, dx`.

`\displaystyle \int_0^\infty (x^(\frac\pi5-1))/(1+x^(2\pi)) \, dx = \frac5\pi \int_0^\infty (dy)/(1+y^(10))`

Now substitute
`y=\left(\frac1z-1\right)^(1/10)` and
`dy=-\frac1{10z^2} \left(\frac1z-1\right)^(-9/10) \, dz` to get a beta integral.

`\displaystyle \int_0^\infty (x^(\pi/5-1))/(1+x^(2\pi)) \, dx = \frac1{2\pi} \int_0^1 z^(-1/10) (1-z)^(-9/10) \, dz \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \boxed{\frac1{2\pi} \, \mathrm{B}\left(\frac9{10}, \frac1{10}\right)}`

We can do better:

Recall that

`\mathrm{B}(a,b) = (\Gamma(a) \Gamma(b))/(\Gamma(a+b))`

as well as the reflection formula for the gamma function,

`\Gamma(z) \Gamma(1-z) = (\pi)/(\sin(\pi z))`

It follows that

`\displaystyle \int_0^\infty (x^(\pi/5-1))/(1+x^(2\pi)) \, dx = \frac1{2\pi} \cdot \frac{\pi}{\sin\left(\frac\pi{10}\right) \Gamma(1)} = \boxed{\frac12 \csc\left(\frac\pi{10}\right)}`

Even better:

To find an exact value for this result, recall

`\sin^2(\theta) = \frac{1-\cos(2\theta)}2`

and

`\cos(5\theta) = 5\cos(\theta) - 20 \cos^3(\theta) + 16 \cos^5(\theta)`

Then

`\theta=\pi \implies -1 = 5\cos\left(\frac\pi5\right) - 20 \cos^3\left(\frac\pi5\right) + 16 \cos^5\left(\frac\pi5\right)`

Let
`t=\cos\left(\frac\pi5\right)`. Solve for
`t` in the quintic equation.

`16t^5 - 20t^3 + 5t + 1 = 0 \\\\ 16 (t^5 - t^3) - 4(t^3 - t) + t+1 = 0 \\\\ (t+1) \left(16t^3 (t-1) + 4t(t-1) + 1\right) = 0 \\\\ (t+1) \left(16t^4 - 16t^3 - 4t^2 + 4t + 1\right) = 0 \\\\ (t + 1) \left(4t^2 - 2t - 1\right)^2 = 0`

Clearly
`t\\eq-1`, so we're left with

`4t^2 - 2t - 1 = 0 \implies t = \frac{1 \pm \sqrt5}4`

and
`t>0`, so we take the positive root.

Now

`\sin^2\left(\frac\pi{10}\right) = \frac{1 - \frac{1+\sqrt5}4}2 = \frac{3-\sqrt5}8 \\\\ ~~~~ \implies \sin\left(\frac\pi{10}\right) = \frac12 \sqrt{\frac{3-\sqrt5}2}`

Denest the radical. Suppose there are rational
`a,b,c` such that

`\sqrt{\frac{3-\sqrt5}2} = a + b\sqrt c`

Squaring both sides gives

`\frac{3-\sqrt5}2 = a^2 + b^2c + 2ab\sqrt c`

Let
`c=5`. Solve for
`a,b`.

`\begin{cases} 2a^2+10b^2 = 3 \\ 4ab = 1 \end{cases} \implies b = -\frac1{4a}`

`a^2 + \frac5{16a^2} = \frac32 \implies 16a^4 - 24a^2 + 5 = 0 \\\\ ~~~~ \implies (4a^2 - 5) (4a^2 - 1) = 0 \\\\ ~~~~ \implies a^2 = \frac54 \text{ or } a^2 = \frac14`

The first case leads to irrational
`a`, so we must have one of
`a=\pm\frac12` and
`b=\mp\frac12`. The value of
`\sqrt{\frac{3-\sqrt5}2}` must be positive, which is consistent with
`a=-\frac12` and
`b=\frac12`.

So we have

`\sin\left(\frac\pi{10}\right) = \frac{\sqrt5-1}4`

and the value of our integral is

`\displaystyle \int_0^\infty (x^(\pi/5-1))/(1+x^(2\pi)) \, dx = \frac12 \cdot \frac4{\sqrt5-1} = \frac{\sqrt5+1}2 = \boxed{\phi}`

(i.e. the golden ratio)